This is the sequel of a previous question.

Let us consider the sequence $$ \xi_n = 2n \{n\xi\}-n, $$ where $\xi>0$ is a given real irrational number and $\{\cdot\}$ is the fractional part.

Do there exist converging subsequences of $\{\xi_n\}_n$ to finite real numbers?

According to an answer, a necessary condition for the existence of cluster points is satisfied: the normalized error $$ q^2 \left| \xi - \frac{p}{q} \right| $$ is bounded with respect to $p$, $q \in \mathbb{N}$ and contains infinitely many different terms. However, it does not seem clear that the convergence of the normalized error along a subsequence entails the convergence of a subsequence of $\xi_n$, because $p$ and $q$ are not arbitrary integers. As far as I understand, diophantine approximation theory is not an obstruction. But is the answer affirmative? It is affirmative only for some class of irrational numbers?

More precisely, we have to produce (if any) a real number $x$ such that, for every $\epsilon>0$ we have for infinitely many integers $n$ $$ \left| n^2 \left(\xi - \frac{[n\xi]}{n}\right) - \frac{x+n}{2} \right| <\frac{\epsilon}{2}. $$


An experimental result.

Take $\xi=\sqrt{2}$.

At least up to $n=706248$, I got subsequence of $8$ terms converging roughly to $C= -1.59\ldots$.

The subsequence coincides so far with OEIS A106328 Numbers j such that 8*(j^2) + 9 = k^2 for some positive number k.

Their $a(n)$ corresponds to $n$ in $\xi_n$ and $\sqrt{2}$ is in the comments.

Convergence also holds for $a(100)$ (76 decimal digits) from OEIS.

With high precision $C= -9\sqrt{2}/8$.

pari/gp session, bugs will invalidate the answer:

? \p 10000
realprecision = 10018 significant digits (10000 digits displayed)
? phi=sqrt(2.0);for(n=1,706248,a=2*n*frac(n*phi)-n;if(abs(round(a))==2,print1(n,",",)));

? phi=sqrt(2.0);for(n=1,706248,a=2*n*frac(n*phi)-n;if(abs(round(a))==2,print([n,precision(a,1)])));
 [3, -1.5441558772842891216]
 [18, -1.5896115822344083765]
 [105, -1.5909496732541739228]
 [612, -1.5909890629760832406]
 [3567, -1.5909902225012345007]
 [20790, -1.5909902566344680254]
 [121173, -1.5909902576392566059]
 [706248, -1.5909902576688348202]
  • $\begingroup$ Even more simply: 3, 18, 105, 612, 3567, 20790, 121173, 706248 are divisible by 3; Dividing by 3 yields 1, 6, 35, 204, 1189, 6930, 40391, 235416 = OEIS A001109 = solutions of $8x^2+1=\Box$, and square roots of square-triangular numbers. (Naturally the A106328 comments include this connection with A00109.) $\endgroup$ – Noam D. Elkies Apr 24 '16 at 17:58

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